I want to show that every $n$ topological manifold $M$ is an $n$ Manifold with boundary where $\partial M=\emptyset$. i.e. every chart $(U,\phi)$ maps to an open set $V\subseteq\mathbb{H}^{n\circ}$ (interior).

If $(U,\phi)$ is a chart on $M$ and $\phi(U)\subseteq\mathbb{R}^n$, then by translation $T$, we can move $\phi(U)$ into the interior of $\mathbb{H}^n$. This would show that every topological manifold is a manifold with boundary where $\partial\mathbb{H}^n=\emptyset$.

But I'm a little unsure about something: For any translation $T:\mathbb{R}^n\to\mathbb{R}^n$ that moves $\phi(U)$ into $\mathbb{H}^n$, why is it necessarily true that $T\circ\phi(U)\cap\partial\mathbb{H}^n=\emptyset$?

My thoughts are since $T:\mathbb{R}^n\to\mathbb{R}^n$ is a homeomorphism, $T\circ\phi(U)\cap\partial\mathbb{H}^n$ is an open subset of $\mathbb{R}^n$. But since $T\circ\phi(U)$ lies entirely in $\mathbb{H}^n$, if $T\circ\phi(U)\cap\partial\mathbb(H)^n\neq \emptyset$ then $T\circ\phi(U)$ would not be open in $\mathbb{R}^n$. Contradiction. Is this correct?


Assuming you have found a translation $T$ such that $T\circ\phi(U) \subseteq \Bbb H^n$, your argument is correct. But it is not really necessary, because if $T\circ\phi(U)$ interected the boundary, just choosing a different translation that moved $\phi(U)$ a little further would also guarantee no intersection.

But, there is no guarantee such a translation exists in the first place. $\phi(U)$ could even be all of $\Bbb R^n$. You are on the right track, but translations are not sufficient by themselves.


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